# Question Video: Finding the Direction Vector of a Line Given Its Slope Mathematics

Which of the following is the direction vector of the straight line whose slope is −1/2? [A] ⟨−1, 2⟩ [B] ⟨1, 2⟩ [C] ⟨2, −4⟩ [D] ⟨2, −1⟩ [E] All the answers are correct.

02:52

### Video Transcript

Which of the following is the direction vector of the straight line whose slope is negative one-half. Is it (A) negative one, two? Is it (B) one, two? Is it (C) two, negative four? (D) Two, negative one. Or (E) all the answers are correct.

Let’s begin by thinking about how to represent a straight line in vector form. The vector form of a straight line is 𝐫 equals 𝐫 sub zero plus 𝑡 times 𝐝. 𝐫 sub zero is a position vector of a point that lies on the line, 𝑡 is a scalar, and then 𝐝 is the direction vector.

Let’s look at this graphically. Suppose we have a point on our line with coordinates 𝑟 sub zero. The position vector 𝐫 sub zero takes us from the origin to that point on our line. Then the direction vector indicates the direction in which the line is moving. The scalar tells us that this can extend in any direction for any length of time. And so it follows that the direction vector would indeed be closely linked to the slope. And of course, we could imagine up a line who has a slope of negative one-half and look to find a couple of coordinates that lie on that line.

But there is a closer link to the direction vector and the slope of a line. In particular, if we know that a line has slope 𝑚, then that line also has direction vector one 𝑚. Now, of course, this direction vector is multiplied by a scalar. So actually, we can have any multiple of this single vector. We’re told that the slope of our line is negative one-half. So we define 𝑚 to be equal to negative one-half. And we can say that the direction vector of our line could be one, negative one-half. But of course, we said it could be any multiple of this. So we can multiply through by two to give us a direction vector of two, negative one. And then, when we do, we see that is equivalent to option (D).

But let’s just quickly disregard all the other options. We can see straightaway that the vector one, negative one-half cannot be multiplied by any constant scalar to give the vector one, two. So we can disregard option (B). Similarly, when we multiplied our vector by two, we got two, negative one. This cannot ever be equal to the vector two, negative four. And we can equally disregard option (A).

To get from one to negative one, we need to multiply by negative one. So in fact, this would give us the vector negative one, one-half. This is not equal to the vector negative one, two. And so by its very nature, option (E) must also be incorrect. This means then the answer is (D). The direction vector of the straight line whose slope is negative one-half is two, negative one.